Help with applied calculus please!!!?

Question

See if you can figure this one out:

Given f(x)=5-e^(-2x),

Determine the equation of the tangent line to the graph of x when x=0.

Any takers?

Thanks in advance!

Answer 1

First you need to take the derivative.

f'(x)= 2*e^(-2x)

Now plug in zero for x.

f'(0)=2*e^(-2(0))
f'(0)=2*e^0
f'(0)=2

So your slope is 2. Now you need to find your y-coordinate.
Plug in zero to the original equation.

f(0)=5-e^(-2(0))
f(0)=5-e^0
f(0)=4

Now you need to plug these three values, x=0, y=4, and m=2, into the point-slope equation.

y-y1=m(x-x1)
y-4=2(x-0)

And your answer is,

y=2x+4

Answer 2

With the slope and the intercept, the tangent can be had. Differentiating to get the slope, we have df/dx = 2 exp(-2x), which at x = 0 is 2. Also, f(0) = 4. So, we have 4 = 2x + c, so c = 4 and the final result is y = 2x + 4.

Answer 3

The tangent line is simply the derivative of the function. Once you do that just plug in x = 0

f'(x) = 2e^(-2x), at x = 0, you get =>f'(0) = 2e^(-2(0)) = 2e^0 =2(1) =2.

Answer 4

f(0)=5-1=4 for obvious reasons. So on f(x) we have point (0,4)

at x=0, slope m= f prime (x)= f p(x) (notation for 1st der I just made up).
f p(x)=2e^(-2x) f p(0)=2=

point-slope format is now y-4=2(x-0) for the tangent, as that line has the slope of 2 and the point (0,4).

Answer 5

Let -2x= u
f(u) = 5-e^u
df(u)/du * du/dx = f'(x)
So du/dx = -2
df(u)/du = -e^u
and f'(x) = 2e^(-2x)
A tangent line has the same slope as f'(x) at x=0
f'(0) =2